Algebra Solutions: Hungerford I.4

1) Let G be a group and \{H_i \mid i\in I\} a family of subgroups. Then for any a\in G \text{, } (\cap_i H_i)a = \cap_i H_ia.
Proof-
(\cap_i H_i)a = \{x\in G \mid x \equiv_r a \pmod{\cap_i H_i}\} = \{ x\in G \mid xa^{-1} \in \cap_i H_i \}= \{ x\in G \mid xa^{-1} \in H_i \forall i \} = \{ x\in G \mid x \equiv_r a \pmod{H_i} \forall i \} = \{ x\in G \mid x \in H_ia \forall i \} = \cap_i H_ia

3) The following conditions on a finite group G are equivalent.
(i) |G| is prime
(ii) G \neq \langle e \rangle and G has no proper subgroups.
(iii) G \cong \mathbb{Z}_p for some prime p.
Proof-
(i) \Rightarrow (iii) Assume that the order of G is a prime, p. Then G has at least 2 elements. Pick an element a, which is not the identity and consider \langle a \rangle. By Lagrange, we know the order of a divides the order of the group. Thus the order of a is either 1 or p, but it is not 1 since a is not the identity. Thus a generates G, so G is cyclic of order p and hence isomorphic to \mathbb{Z}_p since any finite cyclic group of a certain order is actually the same.
(iii) \Rightarrow (ii) Assume that G \cong \mathbb{Z}_p for some prime p. Then the order of G is p which is greater than or equal to 2, so G is nontrivial. Let H \leq G. By Lagrange again, the order of H divides the order of G, so that the order of H is either 1 or p. The first case gives H to be trivial, while the second gives H to be the entire group. Thus G has no proper subgroups.
(ii) \Rightarrow (iii) Assume that G is not trivial and G has no proper subgroups. Since G is nontrivial, there is some a in G such that a is not the identity. Since G has no proper subgroups, \langle a \rangle = G so that G is cyclic and of finite order, hence isomorphic to \mathbb{Z}_m for some integer m. Suppose m is not prime then there is some natural number k such that k divides m and k is not 1 or m. Then because G is cyclic and k divides the order of G, there is exactly 1 subgroup of order k. However we know G has no proper subgroups, so this is a contradiction. Therefore m is prime.
(iii) \Rightarrow (i) Trivial

4) Let a \in \mathbb{Z} and p be a prime such that p \nmid a. Then a^{p-1} \equiv 1 \pmod{p}.
Proof-
By the division algorithm a = pq + r for some q,r \in \mathbb{Z} with 0 < r < p. Thus taking the mod of both sides by p we get a \equiv r \pmod{p}. Recall that the nonzero elements of \mathbb{Z}_p form a group of order p-1 under multiplication mod p. Since 0<r<p r is in this group. Therefore the order of r, call it n, must divide p-1, the order of the group. Thus p-1 = nm for some integer m. So we have:
a^{p-1} \equiv r^{p-1} \pmod{p} but
r^{p-1} =  r^{nm} = (r^{n})^m = 1^m = 1 so that
a^{p-1} \equiv 1 \pmod{p}

6) Let H, K be subgroups of a group G. Then HK \leq H \text{ iff } HK = KH
Proof-
( \Rightarrow ) Assume that HK \leq G. We prove that HK = KH by subset inclusion. Let g\in HK Then g can be written g = hk for some h \in H \text{, } k \in K. Since HK \leq G \text{ then } g^{-1} \in HK as well. Note, g^{-1} = (hk)^{-1} = k^{-1}h^{-1} so that we know k^{-1}h^{-1} \in HK. This tells us that there is some h' \in H \text{ and } k' \in K such that k^{-1}h^{-1} = k'h'. Since H and K are subgroups of G, (h')^{-1} \in H \text{ and } (k')^{-1} \in K. Putting everything together we have that:
g = (g^{-1})^{-1} = (k^{-1}h^{-1})^{-1} = (k'h')^{-1} = (h')^{-1}(k')^{-1} \in KH.
Now let g \in KH so that g=kh for some k \in K \text{ and } h \in H. Then we have that g^{-1} = (kh)^{-1} = h^{-1} k^{-1}. Since H and K are subgroups we have h^{-1} \in H and k^{-1} \in K, so that g^{-1} = h^{-1}k^{-1} \in HK and HK is a subgroup so that g = (g^{-1})^{-1} \in HK also. Thus HK = KH
( \Leftarrow ) Assume now that HK = KH and show HK \leq G. Since H and K are subgroups of G, we have that the e \in H \cap K so that e = ee \in HK which makes HK nonempty. Now we use the subgroup test:
Let hk,h'k' \in HK Then (hk)(h'k')^{-1} = hk(k')^{-1}(h')^{-1} Note that k(k')^{-1}(h')^{-1} \in KH = HK so that we can write k(k')^{-1}(h')^{-1} = h''k'' Then we have:
(hk)(h'k')^{-1} = hh''k'' \in HK. So HK is a subgroup of G.

7) Let G be a group s.t. |G| = p^km where p is prime and gcd(p,m) = 1. Let H, K be supgroups of G s.t. |H| = p^k \text{ and } |K| = p^d with 0 < d \leq k and K not a subgroup of H. Show that HK \nleq G
Proof-
(By contradiction) Assume that HK \leq G then we get |HK| = \frac{|H||K|}{|H \cap K|} = \frac{p^kp^d}{|H \cap K|}. We know that the intersection of subgroups is a subgroup so H \cap K \leq G so that in order for |HK| \in  \mathbb{N} it must be the case that |H\cap k| = p^s for some s \leq k+d. However since K is not a subgroup of H, the intersection of H and K must contain strictly less elements than K. Thus:
p^s = |H\cap K| < |K| = p^d so that s < d. Now we have that |HK| = p^{k+d-s}. By Lagrange the order of the subgroup must divide the order of G, so that p^{k+d-s} \mid p^km which is the same as saying, for some n \in \mathbb{N} that:
p^km = p^{k+d-s}n \Rightarrow m = p^{d-s}n \Rightarrow m = p(p^{d-s-1}n) \Rightarrow p \mid m This is a contradiction because gcd(p,m) = 1. Thus HK is not a subgroup of G.

8) If H and K are subgroups of finite index of a group G such that [G:H] \text{ and } [G:K] are relatively prime, then G = HK.
Proof-
First note two facts: H\cap K \leq H \leq G and H\cap K \leq K \leq G so by theorem 4.5 we can say:
[G: H\cap K] = [G: K] [K: H\cap K] and
[G: H\cap K] = [G: H] [H: H\cap K]. This tells us that [G: H\cap K] is a common multiple of both [G:H] \text{ and } [G:K]. However because gcd([G:H],[G:K]) = 1 we have that lcm([G:H],[G:K]) = [G:H][G:K]. Then clearly because [G: H\cap K] is a common multiple it must be larger than or equal to the least common multiple so that [G: H\cap K] \geq [G:H][G:K]. Lastly, by proposition 4.9 we have [G: H\cap K] \leq [G:H][G:K]. Therefore [G: H\cap K] = [G:H][G:K] so again by proposition 4.9 we can say that G = HK.

9) If H, K, and N are subgroups of a group G such that H\leq N, then HK \cap N = H ( K\cap N)
Proof-
By subset inclusion. Let x\in HK \cap N Then we have x\in HK \text{ and } x\in N. From this we can say x= hk for some h\in H , k\in K. So k = h^{-1}x Since H is given as a subgroup of N this means h^{-1} \in N. so that k = h^{-1}x \in N by closure. But k\in K so that k\in K \cap N Thus:
x = hk\in H(K\cap N) since h\in H, k \in K cap N.
Now let x \in H(K\cap N) This says that x =hg where h\in H \text{ and } g\in K\cap N. Since this tells us g is in K we have x\in HK. Note H is a subgroup of N so that h is in N, and also g is in N. Thus their product is in N. So in conclusion we have x\in HK \cap N

10) Let H,K,N be subgroups of a group G such that H \leq K, H\cap N = K\cap N, \text{ and } HN =KN. Show that H = K.
Proof-
H is already a subgroup of K so we need only show that K is inside of H. Let k\in K. Then k can be written as k = ke \in KN = HN which says that for some h\in H \text{ and } n\in N we have k = hn. Then h^{-1}k = n \in N. Also we have h^{-1} \in K because H is a subgroup of K. Therefore we have h^{-1}k \in N \text{ and } h^{-1}k \in K so thath^{-1}k \in N\cap K = H\cap N.Note the intersection of H and N is inside of H, so that h^{-1}k \in H. Thus h^{-1}k  = h' \Rightarrow k = hh' \in H. So H = K.

11) Let G be a group of order 2n, then G contains an element of order 2. If n is odd and G is abelian, there is only one element of order 2.
Proof-
Let the order of G be 2n. Consider the set \{e, a_1,a_1^{-1}, a_2,a_2^{-1}, ..., a_m,a_m^{-1}\}. If there does not exist an element a_i such that a_i^{-1} = a_i then the order of this set is 2m+1, which is a contradiction so there must exist at least one element of order 2.
If G is abelian and n is odd, suppose we have two elements a,b \in G such that a^2 = e = b^2. G is abelian so that (ab)^2 = e \Rightarrow \{e,a,b,ab\} is a subgroup of G. But because n is odd, we know that 4 does not divide 2n, so this subgroup cannot exist by Lagrange. Thus there can be only one element of order 2.

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